package org.lep.leetcode.singlenumber;

/**
 * Source : https://oj.leetcode.com/problems/single-number/
 *
 * Created by lverpeng on 2017/9/3.
 *
 *
 * Given an array of integers, every element appears twice except for one. Find that single one.
 *
 * Note:
 * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
 *
 */
public class SingleNumber {

    /**
     * 数组中每个元素都有相同的两个，只有其中一个元素是只有一个的，找出这个元素
     * 1. 对数组排序，将arr[i] 和arr[i-1]、arr[i+1]进行比较，如果又都不相等，那么就是要找的元素
     * 2. 使用hash表，循环数组，判断hash表中是否有这个元素，如果有则删除hash表的元素，否则将当前元素加入hash表，最后hash表中剩下的是要找的元素
     * 3. 使用异或运算，xor，一个数和自身异或运算为0，x^x = 0, x^0 = x
     *
     * @param arr
     * @return
     */
    public int singleNumber (int[] arr) {
        int res = 0;
        for (int i = 0; i < arr.length; i++) {
            res ^= arr[i];
        }
        return res;
    }

    public static void main(String[] args) {
        SingleNumber singleNumber = new SingleNumber();
        System.out.println(singleNumber.singleNumber(new int[]{1,1,2,2,3}) + " == 3");
    }
}
